Question: Denis, Vera, and Yash are rock climbers. Yash is connected to Vera by a $30\,\text{m}$ rope, which is taut, and Vera is connected to Denis by a $25\,\text{m}$ rope, which is also taut. Yash has already completed the climb, and he takes out a camera to take a picture of his friends. If Denis sees an angle of $41^\circ$ between Vera and Yash, what field of view would Yash's camera need to fit both of his friends in the same shot? Do not round during your calculations. Round your final answer to the nearest degree.
Converting the problem into geometrical terms Our problem can be modeled by the following triangle $\triangle ABC$, where we want to find $\angle C=\theta$. $A$ $B$ $C$ $\theta$ $41^\circ$ $30\text{ m}$ $25\text{ m}\;\;\;\;\;\;\;\;\;\;\;\;$ Since we are given two side lengths, we can use the law of sines. When using the law of sines we have to keep in mind the ambiguous case, where the angle can be either acute or obtuse. In our case, as $25<30$, we know $\theta$ must be less than $41^\circ$, so the ambiguous case doesn't apply. Using the law of sines $\begin{aligned} \dfrac{\sin(C)}{AB}&=\dfrac{\sin(A)}{BC}\\\\ \dfrac{\sin(\theta)}{25} &= \dfrac{\sin(41^\circ)}{30} \gray{\text{Substitute}}\\\\ \sin(\theta) &= \dfrac{25 \cdot \sin(41^\circ) }{30}\\\\ \theta &= \sin^{-1}\left(\dfrac{25 \cdot \sin(41^\circ) }{30}\right) \\\\ \theta &\approx 33^\circ \end{aligned}$ The answer Yash's camera would need an angle of $33^\circ$ to fit both friends in the same shot.